Category: Mensuration

Question Number 123018 by ajfour last updated on 21/Nov/20 Commented by ajfour last updated on 21/Nov/20 $${If}\:{rectangles}\:{ABCD}\:{and}\:{PQRS} \\ $$$${are}\:{congruent},\:{find}\:\theta,\:{in}\:{terms}\:{of}\:\boldsymbol{\alpha}. \\ $$ Commented by mr W…

Question Number 188520 by Rupesh123 last updated on 02/Mar/23 Answered by HeferH last updated on 02/Mar/23 Commented by HeferH last updated on 02/Mar/23 $${B}+{R}=\:\left(\frac{{r}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\…

Question Number 188443 by Rupesh123 last updated on 01/Mar/23 Answered by HeferH last updated on 04/Mar/23 $${x}=\mathrm{30}° \\ $$ Commented by HeferH last updated on…

Question Number 188378 by Mingma last updated on 28/Feb/23 Answered by HeferH last updated on 09/Mar/23 $${let}\:{x}\:{be}\:{the}\:{side}\:{of}\:{the}\:{red}\:{triangle}\:{and}\:{y}\:{be}\:{the} \\ $$$${side}\:{of}\:{the}\:{blue}\:{one} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{18}°\right)\:=\:\frac{{y}}{{x}}\: \\ $$$$\frac{{A}_{{r}} }{{A}_{{b}} }\:=\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{18}°\right)}\right)^{\mathrm{2}}…

Question Number 188250 by Rupesh123 last updated on 27/Feb/23 Answered by HeferH last updated on 27/Feb/23 $$\left(\mathrm{1}\:−\:\frac{{h}}{\mathrm{2}}\right)\frac{{b}}{\mathrm{2}}\:=\:{b}\centerdot{h} \\ $$$$\:\frac{{b}}{\mathrm{2}}−\frac{{bh}}{\mathrm{4}}\:=\:{bh} \\ $$$$\:\mathrm{2}\:=\:\mathrm{5}{h} \\ $$$$\:{h}\:=\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$…